Question

Two conducting spheres with diameters of 0.400 m and 1.00 m are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to 7.00 % C. (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be V = 0 at r = 0

Answer 1

Two conducting spheres connected by a wire will share charge until they reach electrostatic equilibrium, with the electric potential being the same for both. The charge is distributed based on sphere size, and the total system potential can be calculated from the charge and sphere radius. To find the equivalent number of electrons after charge distribution, divide the charge on each sphere by the elementary charge value.

Step-by-step explanation:

When two conducting spheres of different sizes are connected by a thin wire and charged to a certain quantity, they reach an equilibrium where the electric potential of both spheres is the same. Given the diameters of 0.400 m and 1.00 m, these spheres still operate under the principle that the electric potential on each sphere will become equal once they are connected by a wire. The potential of the spheres will be equivalent due to the movement of electrons through the connecting wire until potential difference ceases (each sphere reaching the same potential).

In this case, the charge distributes in such a manner that the surface charge density (charge per unit area) is the same on both spheres because they are conductors in electrostatic equilibrium. However, since the spheres have different sizes, they will not end up with the same amount of charge. Instead, the larger sphere will have a larger amount of charge due to its greater surface area. After the charge is evenly shared and the spheres are separated, we can say that the charge on each sphere is proportional to its radius.

Regarding the potential of the system, it is given by V = kQ/R, where k is Coulomb's constant, Q is the charge on the sphere, and R is the radius of the sphere. Since the system is assumed to be of the same potential across, we would calculate the potential using the charge and radius of either sphere, because, at equilibrium, the potential across both spheres is the same.

As for the calculation of the number of electrons equivalent to the resulting charge on each sphere, we use the elementary charge value and divide the total charge on each sphere by this value (e = 1.602 x 10⁻¹⁹ C).

Answer 2

Part a)

`Q_1 = 2\mu C`

Part b)

`\Delta V = 4.5 * 10^4 V`

Step-by-step explanation:

As we know that conducting sphere is treated like spherical capacitor

so we can say

`(C_1)/(C_2) = (0.4)/(1)`

also we know that

`Q = CV`

so charge on two spheres will be in the ratio of their capacitance

Part a)

`Q_1 + Q_2 = 7 \mu C`

`(Q_1)/(Q_2)= 0.4`

`1.4Q_2 = 7\mu C`

`Q_2 = 5 \mu C`

`Q_1 = 2\mu C`

Part b)

Now potential difference between two sphere can be given as

`\Delta V = (Q)/(C)`

`\Delta V = (2\mu C)/(4\pi \epsilon_0 R_1)`

`\Delta V = (2\mu C)/(4\pi \epsilon_0(0.400))`

`\Delta V = 4.5 * 10^4 V`